Another limit

Try the same argument with
\lim_{x\to\infty}\left(x\ln(x+1)-x\ln x\right)
[1]

@gallardo

When I said "try using the same argument .... ", I was referring to your idea in #2 where you just neglected 1. If you do the same in #4 you should have got zero for that one as well. However, the limit, as you stated correctly, is 1. So neglecting 1 is not a good idea always.

Actually , I forgot to add " won't " in my statement - i. e ,

" That argument won't work here .... "

Wish I could hide this, but latex images show up:

By MVT, \sin \sqrt{x+1} - \sin \sqrt x = \cos \sqrt c \ \times \ \frac{1}{2\sqrt c} for some c in (x,x+1)

Hence,| \sin \sqrt{x+1} - \sin \sqrt x| \le \frac{1}{2\sqrt c} < \frac{1}{2\sqrt x}

That means the function f(x) = \sin \sqrt{x+1} - \sin \sqrt x may be made as small as we please by increasing x . In other words, by continuity it tends to zero as x becomes large.

That's exactly what I had in mind. Basically, we may use the Squeeze theorem after applying LMVT
\sin\sqrt{x+1}-\sin\sqrt{x} = \dfrac{\cos\sqrt{c}}{2\sqrt{c}}
for some c in (x, x+1).
Since cos√c lies between -1 or 1. So we get
-\dfrac{1}{2\sqrt{c}}\leq \sin\sqrt{x+1}-\sin\sqrt{x} \le \dfrac{1}{2\sqrt{c}}, \quad c\in(x,\,x+1)
As x →∞ so does c. And as c→∞, 1/2√c → 0. Hence, by Squeeze theorem, the given limit must be zero.